Unclear about how you can have high RPM but no power

Question

I did see this question: Manifold pressure vs power vs rpm

But I'm still a bit unclear about the following. I've only flown cherokee pa28s, so I might not have empirical knowledge of the following.

I have a throttle lever in the plane which if I push up, the RPM goes up and it seems like propeller spins faster and the plane has more power, like when we climb out during takeoff.

But I was recently watching this video about Bonanza 6861Q which crashed. He said he had full manifold pressure and full RPM but no power. What does that even mean?

In the archer, I've never heard any of my CFIs talk about the manifold pressure. We check oil pressure, ammeter, and the vacuum gauge. Does manifold pressure refer to one of those things? If not, then what is the manifold pressure?
The RPM and power seem synonymous when I fly in my archer. I push up on the throttle, RPM goes up, feels like the plane has more power. Is RPM different from power?
If so, does the throttle increase the power or the RPM?

If your prop falls off you’ll get higher than usual RPM and no power! ;-) — MD88Fan, Jan 10 '22 at 00:43
I believe that when the pilot (video) says he had full manifold pressure and full rpm but no power he was talking about the moment before he lost power. The loss of power (from the NTSB report being read) was a result of some polyester material in the fuel line. So, it appears that when he pushed the throttle forward the fuel line was essentially clogged and there was no power available because the engine was in the process of failing. Full manifold press and full RPM would not be present if no power was being developed. — , Jan 10 '22 at 01:01
@CaiusJard, well, if you don't have power, you can't have thrust either, but with propeller engines usually power, not thrust, is the characteristic variable we talk about. And power delivered to the propeller at that, not all of which gets converted to thrust. — Jan Hudec, Jan 11 '22 at 15:40

Romeo_4808N · Accepted Answer · 2022-01-11T17:21:39.307

15

The PA-28 Archer uses a fixed pitch propeller. Typically an aircraft with such a propeller, they are not fitted with a manifold pressure gauge as a means for gauging power output. Power output on a fixed pitch propeller is gauged by the engine tachometer. A Beech Bonanza is equipped with a constant speed propeller and will also be equipped with a manifold pressure gauge. This gauge will be used as a measure of power output for the engine.

You really won’t get into the use of manifold pressure until you start flying complex airplanes which have a controllable pitch propeller.

It’s not that a PA-28 engine does not have manifold pressure associated with it. It does. But there is no gauge mounted in the aircraft to measure it.

Neither manifold pressure, nor engine speed are exacting indications of power output; they are sloppy indications and can just be used as a general yardstick to approximate how much power an engine is producing, similar to the torque gauge in a turbopropeller airplane or an N1 or EPR gauges for measuring thrust output from a jet. In fact if you look at the cruise performance charts for a complex airplane, you’ll know that a particular power output requires both a manifold pressure and propeller speed setting in order to ensure an exact power output from the engine. Similarly in an airplane with a fixed pitch propeller they will do a general gauging of power based on propeller speed in the cruise performance charts. Note this power varies with altitude i.e. you will not get the same power out of the engine at 2400 RPM while flying at 6000 feet, versus flying at 3000 feet.

Remember that in an engine manifold pressure describes in fact the pressure of the fuel air mixture entering the engine manifold. When the engine is not operating, the manifold pressure will return to whatever the ambient atmospheric pressure is. Should an engine failure occur, the manifold pressure will not change, and will remain constant with a throttle setting provided the engine continues to turn. Typically manifold pressure is measured in inches of mercury, so for that bonanza if it’s sitting on the ground, engine off, at sea level @ STP, the manifold pressure gauge would read 29.92 inches of mercury.

Neither manifold pressure nor engine RPM are a reliable indication of an engine failure and should never be used as such for the reasons stated above. That was the problem the pilot was perceiving when he lost engine power. As soon as the engine power went the manifold pressure gauge returned to ambient atmospheric pressure, which coincidentally is exactly the same as if the engine was operating with a throttle wide open. Propellers can often windmill at maximum speed in a dive, which also prevents them from being a reliable indication of engine power. Your two most reliable gauges for determining engine failure in an AvGas powered piston airplane are 1) the fuel flow gauge and 2) the exhaust gas temperature (EGT) gauge or turbine inlet temperature gauge (TIT) for a turbocharged airplane. That seems counterintuitive, but for the problems I listed above with manifold pressure gauges it’s the only reliable way to determine whether engine is properly operating.

The throttle in a light training airplane like a PA-28 only controls the butterfly throttle valve in a carbureted airplane, or the fuel/air control unit in a fuel injected airplane. Engine tachometer only measures the rotational speed of the driveshaft and propeller

edited Jan 11 '22 at 17:21

answered Jan 09 '22 at 20:20

Romeo_4808N

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“As soon as the engine power went the manifold pressure gauge returned to ambient atmospheric pressure” – no, the manifold pressure really remained unchanged. Manifold pressure depends on how much air the engine sucks in and the throttle position only, and with the prop governor keeping constant RPM the engine is still pulling through the same amount of air when driven by the windmilling propeller. Only on a turbocharged engine the MP would (gradually) drop as the turbo winds down, because the exhaust is no longer expanded through combustion. – Jan Hudec Jan 10 '22 at 14:20
Also if you do have a torque monitor that is an accurate indicator, because power equals torque times angular velocity (RPM), by definition. So if you have those two, you know the power, exactly. For that reason, auto-feather usually uses negative reading on torque as trigger. But I don't think any light piston, even complex, has torque indicator. Only turbines, and some vintage large pistons (like DC-6 or Constellation) do. – Jan Hudec Jan 10 '22 at 14:26
… you can also still have fuel flow and no power if it's ignition that gave out, so that leaves EGT as the only reliable indicator the engine is not working properly. – Jan Hudec Jan 10 '22 at 14:27
… if the engine actually stopped, the manifold pressure would return to ambient, but in this case it didn't stop, it was still windmilling at the constant RPM. – Jan Hudec Jan 10 '22 at 15:54
Well the same could be set for a piston airplane in that if you have a manifold pressure gauge plus a engine tachometer, you can gauge how much power you are producing based on a conversion table. In addition your comment about toric is an accurate, as the displayed engine speed would have to be in rads per second to work without a conversion factor of some sort. – Romeo_4808N Jan 10 '22 at 16:21
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No, you can't, and that's exactly what this accident, and disussion demonstrates. On a normally-aspirated engine when you turn off the ignition, but let the governor keep the RPM of the widmilling propeller constant, neither MP, RPM or fuel flow will move but for a tiny fluctuation as the governor catches up. – Jan Hudec Jan 10 '22 at 17:16
As for units, it does not matter what units you have. If you have torque in $\mathrm{lb_f}·\mathrm{ft}$ and angular speed in revolutions per minute, you get power in $π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}$, which is somewhat weird unit, but it's just a unit and it is still a unit of power (multiply by 1 expressed as $0.071 \frac{\mathrm W}{π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}}$ or $9.5·10^{-5} \frac{\mathrm{hp}}{π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}}$ to get something reasonable). – Jan Hudec Jan 10 '22 at 17:26
I don’t think you’re understanding what I’m saying in my comments Jan. I’m not talking about the event of an engine failure – Romeo_4808N Jan 10 '22 at 18:29
That still remains sloppy, Jan. Well torque and power use the same units, a torque gauge is in fact not indicating your engines power output, and you will have to perform the additional calculations you suggest above. Again it can serve as a general yardstick to use for particular flight situations i.e. on approach at Vref, a/c configured for landing and trimmed neutral the airplane remains stabilized at 500 ft-lbs, max RPM, etc. – Romeo_4808N Jan 10 '22 at 19:05
The whole question is about engine failure, and the difference in response to engine failure was the key point of my comment that you responded to. 2. It isn't sloppy. Torque times angular velocity is power. If you have those two gauges – I never said torque alone – you know the power, and you know it as long as the gauges themselves, independent of anything else, work. Unlike MP, which needs many further assumptions. 3. No, torque and power definitely don't have the same unit, nor do they have the same dimension.

Jan Hudec

Jan 11 '22 at 15:36

"a beach Bonanza" - would it need a sand filter? – Michael Harvey Jan 11 '22 at 15:45

score 9 · Answer 2 · edited Jan 10 '22 at 16:51

From a physics standpoint, power is defined to be (force x velocity); in rotary shaft work terms this is (torque x RPM).

This means that a device that produces 100 ft-lbs of torque at zero RPM (shaft-locked condition) is producing zero power, and a device that produces 2500RPM with the load disconnected (no-load condition) is producing zero power.

If that device produces 100 ft-lbs of torque at 2500RPM, then it is producing (100 x 2500/5252) shaft horsepower where the 5252 divisor converts the units to horsepower.

With a fixed pitch propeller, the relationship between the engine load and the prop RPM is fixed, which means if you know your RPM, you can look up the corresponding horsepower on a chart displaying a power vs. RPM line.

With a variable-pitch propeller, that same chart will contain not one line but a family of lines corresponding to different values of the load, as determined by different values of the propeller pitch- and in this case each line is labelled by the manifold vacuum which has been measured on the engine running at that load level and RPM.

Note that manifold vacuum is a proxy variable for the engine load torque since it is a measure of how fast the engine is pulling fuel/air mixture through the manifold.

In rotational motion, HP = Ft-Lbs * RPM / 5252. The 550 is for converting HP = Ft-Lbs/s / 550, or linear motion. (You could also do radians per second or something and get other constants.) — MichaelS, Jan 10 '22 at 05:07

score 1 · Answer 3 · answered Jan 10 '22 at 14:43

As already mentioned, a Beechcraft Bonanza has a constant speed propeller.

Constant speed propeller has a governor that drives the propeller coarser if it's too fast and finer if it's too slow. When the engine loses power, the propeller is driven to fine where it can maintain the RPM just through windmilling so RPM won't change.
Manifold pressure indicates how the pressure is reduced by the throttle body limiting the amount of air being inducted in the engine. Since at the same RPM the engine still inducts the same volume per second and the throttle body didn't move, the manifold pressure does not move either. It would only move if the problem was inlet obstruction.
In this case, the fuel flow could have been seen dropped as the cause was fuel line clogging, but if it was ignition failure, the fuel would still be inducted too.
Only EGT, exhaust gas temperature, dropping is a reliable indication that there no (or less) combustion going on and therefore the engine is failing. And the feel, of course.

If it was a turbocharged engine, the manifold pressure would drop (with some delay) as the turbo would wind down, because without combustion the volume of exhaust is less and wouldn't be able to drive it any more. But with normally aspirated engine it usually just stays the same in a failure. It is a useful reference for accurately setting the standard cruise power, and to avoid overboosting a turbocharged engine on take-off, but does not help in case of failures.

Unclear about how you can have high RPM but no power

3 Answers3